How to evaluate the stability of a full wave precision rectifier

Question

I made a simulation of a full wave precision rectifier but I have some difficulties to evaluate its stability.

I would like to evaluate it stability by a Bode plot because it seems that if the input of the circuit is negative it starts to oscillate when I do a transient analysis on LTspice.

As the feedback path used is not the same if the input is positive or negative, I was thinking that it was possible to simplify the circuit and obtain two stage of simple op amp circuit. In this way, I was thinking that I could evaluate the stability of each stage and then get the total stability of the system for different operating point. Nevertheless I do not think that it is so simple and I would probably evaluate the stability of a system that is not representative of the one I want to study...

Do you know what would be the correct method for evaluating the stability of such a system on LTspice ?

Version 4
SHEET 1 1268 980
WIRE -592 -32 -768 -32
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SYMBOL Opamps\\OPA810 176 160 R0
SYMATTR InstName U1
SYMBOL res 224 -48 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R1
SYMATTR Value 10k
SYMBOL res 32 128 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R2
SYMATTR Value 10k
SYMBOL voltage -192 176 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value PULSE(1 2 10u 1u 1u 10u)
SYMBOL res 224 304 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R3
SYMATTR Value 10k
SYMBOL Opamps\\OPA810 560 176 R0
SYMATTR InstName U2
SYMBOL res 464 -48 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R4
SYMATTR Value 10k
SYMBOL res 656 48 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R5
SYMATTR Value 10k
SYMBOL res 784 208 R0
SYMATTR InstName R6
SYMATTR Value 100k
SYMBOL voltage -448 -32 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V2
SYMATTR Value 15
SYMBOL voltage -768 -32 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V3
SYMATTR Value -5
SYMBOL Opamps\\OPA810 176 656 R0
SYMATTR InstName U3
SYMBOL res 224 448 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R7
SYMATTR Value 10k
SYMBOL res 32 624 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R8
SYMATTR Value 10k
SYMBOL voltage -192 672 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V4
SYMATTR Value PULSE(-1 -2 10u 1u 1u 10u)
SYMBOL res 224 800 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R9
SYMATTR Value 10k
SYMBOL Opamps\\OPA810 560 672 R0
SYMATTR InstName U4
SYMBOL res 464 448 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R10
SYMATTR Value 10k
SYMBOL res 656 544 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R11
SYMATTR Value 10k
SYMBOL res 784 704 R0
SYMATTR InstName R12
SYMATTR Value 100k
SYMBOL diode 304 512 R0
SYMATTR InstName D3
SYMBOL diode 304 704 R0
SYMATTR InstName D4
SYMBOL diode 304 16 R0
SYMATTR InstName D1
SYMBOL diode 304 208 R0
SYMATTR InstName D2
TEXT -226 370 Left 2 !.tran 30u
TEXT 1032 -8 Left 2 !.MODEL BAS16LD D\n+ IS = 6.0289E-9\n+ N = 2.1283\n+ RS = 1.0000E-3\n+ IKF = 37.638E-3\n+ CJO = 5.641E-013\n+ M = 0.05188 \n+ VJ = 0.01571\n+ ISR = 2.0646e-9\n+ NR = 4.9950\n+ BV = 113.60\n+ IBV = 10\n+ TT = 4.5000E-9\n*##\n*

Thank you

Answer 1

When an AC steady-state analysis is done, a DC .op (operating point) analysis is done first. This determines which bias levels, feedback paths, etc. are active in the circuit. In short, diodes are transformed into resistors or capacitors (some combination), transistors are transformed into those and dependent sources, etc.

To provide this information, simply set a DC parameter for your PULSE source(s), then do the AC analysis. Adjust RC values until the gain is smooth. Usually, instability shows up as a sharp but still finite peak; reducing it to a mildly or non-peaked response should be good enough. Then go back and forth between transient and AC analyses to fine tune the response, getting step overshoot or settling time as desired. (For these transient analyses, check the response to a small step (100mV or less?) at various DC offsets. You may want to increase accuracy (e.g. reduce RELTOL) to get better resolution on these small changes.)

The transient analysis computes the effect of each element at each timestep, so has a more realistic (nonlinear) response, especially near zero where both diodes are switching on/off. But it is harder to do steady state AC analysis with (that said, LTSpice has tools to address these too, now).

Remember to put in realistic component models, and board strays if you can estimate them. At the highest bandwidths (100s MHz?, or less if breadboarded, heh), trace inductances will also become important.

By "RC values", I mean the value of all resistors in the circuit (or some ratios of them, in the case you need gain from this rectifier) with respect to diode junction, op-amp input, or stray node capacitances, or the addition of C or R+C elements in places to reduce gain and phase shift around any given op-amp. For example, in parallel with R5 to reduce U2's loop gain; or with R3; or across U1 directly (OUT to -IN).

A typical issue I have with this circuit is using too large resistors or diodes. For example, BAT54S has relatively high capacitance, and 1N4148 would be better, or BAS70 better still; 10k resistors are probably sufficient for ~3MHz GBW op-amps, but for 20+ MHz you may want to reduce that to 2.2k or less.

As you haven't put in diode or op-amp models, it is impossible for me to give a particular resolution, but a general method was asked so this should suffice.

(As a final note, the asymmetric supplies aren't important, at least in absence of any op-amp model, and not for well-behaved op-amps. Ensure the AC analysis is not done in saturation (op-amp outputs near supply rails). That said, you may also be interested in testing the transient response in this region: i.e. step response into/out of saturation, or recovery from overdrive say with a sine or triangle wave source.)

Answer 2

There is a "pitfall" in this conception...

Here's what it should be. (R6 added because Vd2 is "lost" in some cases).

Answer 3

Here is a precision full wave rectifier circuit from National Semiconductor: